Question 10.6: A transfer function is given by H(s)= 72s/(s+3)(s+6).Sketch ......

Substituting s = jω in the transfer function yields,

$H(\omega)={\frac{72(j\omega)}{(j\omega+3)(j\omega+6)}}$     (10.190)

$H(\omega)=\frac{72(j\omega)}{3(\frac{j\omega}{3}+1)\times6\left(\frac{j\omega}{6}+1\right)}$     (10.191)

$H(\omega)=\frac{4(j\omega)}{(\frac{j\omega}{3}+1)\left(\frac{j\omega}{6}+1\right)}$     (10.192)

The expressions of the magnitude and phase are,

$H_{\mathrm{dB}}=20\log_{10}(4)+20\log_{10}(j\omega)-20\log_{10}\biggl|\frac{j\omega}{3}+1\biggr|-20\log_{10}\biggl|\frac{j\omega}{6}+1\biggr|$     (10.193)

$\phi=90^{\circ}-\tan^{-1}\left({\frac{\omega}{3}}\right)-\tan^{-1}\left({\frac{\omega}{6}}\right)$     (10.194)

The magnitude and phase plots are shown in Figs. 10.34 and 10.35, respectively.